﻿ If x-\dfrac{2}{\sqrt{x} } =5, find the value of x-2\sqrt{x}

#### Question

If x-\dfrac{2}{\sqrt{x} } =5, find the value of x-2\sqrt{x}

Collected in the board: Radical equations

Steven Zheng posted 4 months ago

Given

x-\dfrac{2}{\sqrt{x} } =5
(1)

Let

a = \sqrt{x}
(2)

then

x = a^2
(3)

Equation (1) is transformed to

a^2-\dfrac{2}{a} = 5

then a cubic equation

a^3-5a -2=0
(4)

Let

A =x-2\sqrt{x}
(5)

Substitute x with a

A = a^2-2a
(6)

or

a^2 = A+2a
(7)

Multiply the equation with a

Aa = a^3-2a^2
(8)

Substitute a^2 with (7)

Aa = a^3-2(A+2a)

Rearrange the terms

a^3-(4+A)a-2A = 0
(9)

Compare the cubic equation with (4)

4+A = 5

Then A = -1

Therefore

x-2\sqrt{x} = -1

Steven Zheng posted 4 months ago

x-\dfrac{2}{\sqrt{x} } =5, find the value of x-2\sqrt{x}

x-\dfrac{2}{\sqrt{x} } = 1+4

Rearrange the terms

x-4 = \dfrac{2}{\sqrt{x} } +1

Using difference of squares formula

(\sqrt{x}-2 )(\sqrt{x}+2 ) = \dfrac{2+\sqrt{x} }{\sqrt{x} }

Cancel \sqrt{2} +2 term since it's larger than 1

\sqrt{x}-2 = \dfrac{1}{\sqrt{x} }

(\sqrt{x}-1 )^2 =0

Then x = 1

x-2\sqrt{x} =-1

Steven Zheng posted 4 months ago

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